0
$\begingroup$

the equation is

$\ln(x+2)=\ln e^{\ln2} - \ln x$

How do I solve for $x$?

$\endgroup$
  • 1
    $\begingroup$ $e^{\ln 2} = 2$, for starters $\endgroup$ – The Chaz 2.0 Oct 23 '13 at 4:03
  • $\begingroup$ can you explain why ? $\endgroup$ – user293849 Oct 23 '13 at 4:26
  • $\begingroup$ @user293849: Think about it: What does "$\ln 2$" mean? $\endgroup$ – Blue Oct 23 '13 at 4:27
  • $\begingroup$ @user293849 Look up the logarithm identities $\ln a^b = b \ln a$. In your case $b=\ln 2$ and $\ln e=1$. Please have a look here en.wikipedia.org/wiki/List_of_logarithmic_identities $\endgroup$ – triomphe Oct 23 '13 at 4:32
3
$\begingroup$

Your equation is equivalent to $$\ln (x+2)+\ln(x)=\ln2.$$ Then use logarithm identity $\ln ab=\ln a +\ln b$\$ $$\ln (x+2)x=\ln2.$$ Now take inverse of both sides $$(x+2)x=2.$$ Now you can solve the quadratic equation and select the appropriate $x$ value. You get $x=-1\pm\sqrt 3.$ Note that you can not take $x<0$ as $\ln x$ is not defined for negative $x$. So your answer is $x=-1+\sqrt 3$ which is approximately equal to $0.732$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.